Q56CP - University Physics with Modern Physics

Question

An object of mass $\mathbf{m}$ is at rest in equilibrium at the origin. At $\mathbf{t}\mathbf{=}\mathbf{0}$ a new force $\overrightarrow{\mathbf{F}}\left(t\right)$  is applied that has components ${\mathbf{F}}_{\mathbf{x}}\left(t\right)\mathbf{=}{\mathbf{k}}_{\mathbf{1}}\mathbf{+}{\mathbf{k}}_{\mathbf{2}}\mathbf{y}$ and ${\mathbf{F}}_{\mathbf{y}}\left(t\right)\mathbf{=}{\mathbf{k}}_{\mathbf{3}}\mathbf{t}$ . Where, ${\mathbf{k}}_{\mathbf{1}}$ , ${\mathbf{k}}_{\mathbf{2}}$ , and ${\mathbf{k}}_{\mathbf{3}}$ are constants. Calculate the position $\overrightarrow{\mathbf{r}}\left(t\right)$ and velocity  $\overrightarrow{\mathbf{v}}\left(t\right)$ vectors as functions of time.

Step-by-Step Solution

Answer

The velocity vector as a function of time is $\left(\frac{{\mathrm{k}}_1}{\mathrm{m}}\mathrm{t}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^4}{24{\mathrm{m}}^2}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{k}}_3}{2\mathrm{m}}{\mathrm{t}}^2\right)\hat{\mathrm{j}}$.

The position vector as functions of time is $\left(\frac{{\mathrm{k}}_1}{2\mathrm{m}}{\mathrm{t}}^2+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^5}{120{\mathrm{m}}^2}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{k}}_3}{6\mathrm{m}}{\mathrm{t}}^3\right)\hat{\mathrm{j}}$.

Step 1: Identification of the given data:

The given data can be listed below as,

• The mass of the object is $\mathrm{m}$.
• The force exerted in the $\mathrm{x}$ direction is ${\mathrm{F}}_{\mathrm{x}}\left(\mathrm{t}\right)={\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{y}$.
• The force exerted in the   direction is ${\mathrm{F}}_{\mathrm{y}}\left(\mathrm{t}\right)={\mathrm{k}}_3\mathrm{t}$.

Step 2: Significance of the position vectors:

The position vector is referred to as a straight line that has one fixed and another flexible end. The position vector is used to locate the position of a point particle.

Step 3: Determination of the velocity vector:

The equation of the acceleration vector as a function of time in the $\mathrm{x}$ direction is expressed as:

 ${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_{\mathrm{x}}\left(\mathrm{t}\right)}{\mathrm{m}}$ 

Here, ${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)$ is the acceleration vector as a function of time in the $\mathrm{x}$ direction,  ${\mathrm{F}}_{\mathrm{x}}\left(\mathrm{t}\right)$ is the force exerted in the $\mathrm{x}$ direction and $\mathrm{m}$ is the mass of that object.

 

Substitute the values in the above equation.

 

${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{y}}{\mathrm{m}}$                                                                                                         ..… (1)

 

The equation of the acceleration vector as a function of time in the $\mathrm{y}$ direction is expressed as:

 

 ${\mathrm{a}}_{\mathrm{y}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_{\mathrm{y}}\left(\mathrm{t}\right)}{\mathrm{m}}$ 

 

Here, ${\mathrm{a}}_{\mathrm{y}}\left(\mathrm{t}\right)$ is the acceleration vector as a function of time in the $\mathrm{y}$ direction,  ${\mathrm{F}}_{\mathrm{y}}\left(\mathrm{t}\right)$ is the force exerted in the $\mathrm{y}$ direction and $\mathrm{m}$ is the mass of that object.

 

Substitute the values in the above equation.

 $a_y\left(t\right)=\frac{k_{3}t}{m}$                                                                                                ….. (2)

 

The equation of the final velocity in the $\mathrm{y}$ direction is expressed as:

  ${\mathrm{v}}_{\mathrm{y}}=\mathrm{u}+\underset{0}{\overset{\mathrm{t}}{\int }}{\mathrm{a}}_{\mathrm{y}}\mathrm{dt}$

 

Here, ${\mathrm{v}}_{\mathrm{y}}$ is the final velocity in the $\mathrm{y}$ direction, $\mathrm{u}$ is the initial velocity and ${\mathrm{a}}_{\mathrm{y}}\left(\mathrm{t}\right)$  is the acceleration vector as a function of time in the $\mathrm{y}$ direction.

 

As the object was at rest initially, hence the initial velocity of the object is zero.

 

Substitute 0 for $\mathrm{u}$ and $\frac{{\mathrm{k}}_3\mathrm{t}}{\mathrm{m}}$ for ${\mathrm{a}}_{\mathrm{y}}\left(\mathrm{t}\right)$ in the above equation.

$$\begin{gathered} v_y=0+\underset{0}{\overset{t}{\int }}\frac{k_{3}t}{m}dt \\ =\frac{k_3{t}^2}{2m} \end{gathered}$$ 

 

The equation of the distance along the $y$ direction is expressed as:

$\mathrm{y}={\mathrm{y}}_0+\underset{0}{\overset{\mathrm{t}}{\int }}{\mathrm{v}}_{\mathrm{y}}\mathrm{dt}$  

 

Here, $\mathrm{y}$ is the final and ${\mathrm{y}}_0$ is the initial distance covered by the object.

 

As the object was at rest initially, hence the initial distance covered by the object is zero.

Substitute $0$ for $\mathrm{y}$ and $\frac{{\mathrm{k}}_3{\mathrm{t}}^2}{2\mathrm{m}}$ for ${\mathrm{v}}_{\mathrm{y}}\left(\mathrm{t}\right)$ in the above equation.

 $$\begin{gathered} \mathrm{y}=0+\underset{0}{\overset{\mathrm{t}}{\int }}\frac{{\mathrm{k}}_3{\mathrm{t}}^2}{2\mathrm{m}}\mathrm{dt} \\ =\frac{{\mathrm{k}}_3{\mathrm{t}}^3}{6\mathrm{m}} \end{gathered}$$

Substitute the value of the above equation in the equation (1).

 ${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{{\mathrm{k}}_1}{\mathrm{m}}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3}{6{\mathrm{m}}^2}{\mathrm{t}}^3$

 

The equation of the final velocity in the $\mathrm{x}$ direction is expressed as:

 ${\mathrm{v}}_{\mathrm{x}}=\mathrm{u}+\underset{0}{\overset{\mathrm{t}}{\int }}{\mathrm{a}}_{\mathrm{x}}\mathrm{dt}$

Here, ${\mathrm{v}}_{\mathrm{x}}$ is the final velocity in the $\mathrm{x}$ direction, $\mathrm{u}$ is the initial velocity and ${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)$ is the acceleration vector as a function of time in the $\mathrm{x}$ direction.

 

As the object was at rest initially, hence the initial velocity of the object is zero.

Substitute $0$ for $\mathrm{u}$ and $\frac{{\mathrm{k}}_1}{\mathrm{m}}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3}{6{\mathrm{m}}^2}{\mathrm{t}}^3$ for ${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)$ in the above equation.

 $$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=0+\underset{0}{\overset{\mathrm{t}}{\int }}\frac{{\mathrm{k}}_1}{\mathrm{m}}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3}{6{\mathrm{m}}^2}{\mathrm{t}}^3\mathrm{dt} \\ =\frac{{\mathrm{k}}_1}{\mathrm{m}}\mathrm{t}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^4}{24{\mathrm{m}}^2} \end{gathered}$$

 

The equation of velocity vector is expressed as:

$\overrightarrow{\mathrm{v}}\left(\mathrm{t}\right)={\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}$ 

Here, $\overrightarrow{\mathrm{v}}\left(\mathrm{t}\right)$ is the velocity vector.

 

Substitute the values in the above equation.

$\overrightarrow{\mathrm{v}}\left(\mathrm{t}\right)=\left(\frac{{\mathrm{k}}_1}{\mathrm{m}}\mathrm{t}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^4}{24{\mathrm{m}}^2}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{k}}_3}{2\mathrm{m}}{\mathrm{t}}^2\right)\hat{\mathrm{j}}$ 

 

Thus, the velocity vector as functions of time is $\left(\frac{{\mathrm{k}}_1}{\mathrm{m}}\mathrm{t}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^4}{24{\mathrm{m}}^2}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{k}}_3}{2\mathrm{m}}{\mathrm{t}}^2\right)\hat{\mathrm{j}}$.

Step 4: Determination of the position vector:

The equation of the distance along the $x$ direction is expressed as:

 

 $x=x_0+\underset{0}{\overset{t}{\int }}{v}_x\left(t\right)dt$ 

 

Here, $\mathrm{x}$ is the final and ${\mathrm{x}}_0$ is the initial distance covered by the object along the $\mathrm{x}$ direction.

 

As the object was at rest initially, hence the initial velocity of the object is zero.

 

Substitute the values in the above equation.

 

 $$\begin{gathered} \mathrm{x}=0+\underset{0}{\overset{\mathrm{t}}{\int }}\frac{{\mathrm{k}}_1}{\mathrm{m}}\mathrm{t}+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^4}{24{\mathrm{m}}^2}\mathrm{dt} \\ =\frac{{\mathrm{k}}_1}{2\mathrm{m}}{\mathrm{t}}^2+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^5}{120{\mathrm{m}}^2} \end{gathered}$$

 

The equation of the position vector is expressed as:

 

$\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}$ 

 

Here, $\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)$ is the position vector.

 

Substitute the values in the above equation.

 

  $\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)=\left(\frac{{\mathrm{k}}_1}{2\mathrm{m}}{\mathrm{t}}^2+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^5}{120{\mathrm{m}}^2}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{k}}^3}{6\mathrm{m}}{\mathrm{t}}^3\right)\hat{\mathrm{j}}$

 

Thus, the position vector as functions of time is $\left(\frac{{\mathrm{k}}_1}{2\mathrm{m}}{\mathrm{t}}^2+\frac{{\mathrm{k}}_2{\mathrm{k}}_3{\mathrm{t}}^5}{120{\mathrm{m}}^2}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{k}}^3}{6\mathrm{m}}{\mathrm{t}}^3\right)\hat{\mathrm{j}}$.